14.6: Confidence Interval for y′ at a Given x
- Page ID
- 52062
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)At a fixed \(x\) (that is important to remember) the confidence interval for \(y\) is
\[ y^{\prime} - E < y < y^{\prime} + E \]where
\[ E = t_{\cal{C}} \; s_{\mbox{est}} \sqrt{1+ \frac{1}{n} + \frac{n(x - \overline{x})^{2}}{n(\sum x^{2})-(\sum x)^{2}}} \]where, as usual, \(t_{\cal{C}}\) comes from the t Distribution Table with \(\nu = n-2\).
Example 14.5 : Continuing from Example 14.4 (so you can see how an exam will go), say we want to predict the grade (\(y\)) in terms of a 95\(\%\) confidence interval for the number of absences (\(x\)) equal to 10.
First, find the value predicted from the regression line, which we previously found to be :
\[ y^{\prime} = 102.493 - 3.622 x \]at \(x = 10\). The result is
\[ y^{\prime} = 102.493 - 3.622 (10) = 66.273 \]Furthermore, from the last example, we found
\[ s_{\mbox{est}} = 6.06 \]and, from the completed data table (Example 14.3)
\[ \sum x = 57 \;\;\; \sum x^{2} = 579 \]We still need \(t{\cal{C}}\) and \(\overline{x}\). Using our sums:
\[ \overline{x} = \frac{\sum x}{n} = \frac{57}{7} = 8.143 \]and from t Distribution Table for the 95\(\%\) confidence interval, \(\nu = 7-2 = 5\) we get
\[ t{\cal{C}} = 2.571 \]Now we compute \(E\) :
\[\begin{eqnarray*} E & = & t_{\cal{C}} \; s_{\mbox{est}} \sqrt{1+ \frac{1}{n} + \frac{n(x - \overline{x})^{2}}{n(\sum x^{2})-(\sum x)^{2}}}\\ E & = & (2.571) \; (6.06) \sqrt{1+ \frac{1}{7} + \frac{7(10 - 8.143)^{2}}{7(579)-(52)^{2}}}\\ E & = & 15.58026 \sqrt{1 + 0.1428571 + \frac{24.139}{804}}\\ E & = & 16.77 \end{eqnarray*}\]So
\[\begin{eqnarray*} y^{\prime} - E & < y < & y^{\prime} + E \\ 66.273 - 16.77 & < y < & 66.273 + 16.77 \\ 49.5 & < y < & 83.0 \end{eqnarray*}\]This is the 95\(\%\) confidence interval for predicting the mark of a person who was absent for 10 days.
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Important: \(s_{\mbox{set}}\) is independent of \(x\) but \(E\) is not. So confidence intervals look like :
The reason for this variance of the width of the confidence interval comes from the uncertainty in the slope \(b\). You can make plots like the one above in SPSS.